Length contraction and direction of tavel

[Amr Elsayed]

I didn't actually. I didn't even read it and I don't know why or how, but If I see 1 sec difference between my watch and the other one at back" after I modified it with your step" then a guy on platform will not measure this difference ? what's next ? When I measure light speed I shall take this second into account, but again since it's same person measuring I then don't have to take it into account . I just want difference in time according to me.
I would be grateful If you tell me what exactly I will measure and how as a one guy in front of moving train. and light is striking me from outside not from train in sth like brief scenario

 

[harrylin]

I didn't actually. I didn't even read it and I don't know why or how, but If I see 1 sec difference between my watch and the other one at back" after I modified it with your step" then a guy on platform will not measure this difference ? what's next ? When I measure light speed I shall take this second into account, but again since it's same person measuring I then don't have to take it into account . I just want difference in time according to me.
I would be grateful If you tell me what exactly I will measure and how as a one guy in front of moving train. and light is striking me from outside not from train in sth like brief scenario

Sorry, your questioning exhausted me, and I already gave all calculations and explanations that you need. More useful if you re-read this thread and try to get more out of it.

I don't remember if I put it to your attention, but we simply repeated here what is written and explained at the end of §2 of http://fourmilab.ch/etexts/einstein/specrel/www/ (starting with "We imagine further")

The essential calculations about clock synchronization are primary school stuff; it's therefore not much to do with calculations, it's conceptual understanding. And for a brief moment in time you had it.

For low speeds you can approximate the Lorentz transformations as follows, and it can be useful to sketch the time transformation along the X and X' axes as I suggested in post 20:
x' ≈ x - vt
t' ≈ t - vx/c²

PS: most likely you need just a few days to let the concept of "relativity of simultaneity" sink in. I recall that this was the case for me, when I finally "got" it!

 

[Amr Elsayed]

Thank you for help and care. you have been explaining in 4 pages now. I will search again about simultaneity and synchronized clocks, and I hope it will sink in.
Thanks once again
regards

 

[Stephanus]

Thank you for help and care. you have been explaining in 4 pages now. I will search again about simultaneity and synchronized clocks, and I hope it will sink in.
Thanks once again
regards

I think you should start from the beginning Amr Elsayed. But I don't know either where the "beginning" is.
I have asked about
Twin Paradox,
Twin Paradox and asymmetry
Twin Paradox and symmetry
Lorentz and Doppler
Motion in Space
Universe frame of reference.
Still I haven't reach my destination, But I know, I'm on the right track based on their answers.
Janus clip have helped me much. Two clips actually he gave me.

1. About synchronizing clocks with and without the observer moving.
https://www.physicsforums.com/threads/twin-paradox-and-asymmetry.816455/
You can find his clips somewhere from that thread.
I think, if an amateur like me understand (or at least grasp), perhaps you can understand more.
From his answer, I think I can understand that simultaneity events are affected by relativity.

2. And in the train clip above, I think I know WHY there has to be length contraction. With out this, the signal, from the train point of view, can't reach the observer at the same time. About the HOW, well, still don't understand. Yet. Or, still haven't calculated it yet. I think it somehow has something to do with Lorentz Transformation.
And that's why I create a new thread about Lorentz Transformation before I make some calculations.
And length contraction leads to affected simultaneity events.
Just don't be discourage if you don't understand

 

[Mentz114]

@Amr Elsayed

I think what you are trying to do is like proving that a triangle has three sides. The speed of light is a constant in SR by definition just like a triangle has three sides by definition.
If your numerical calculations tell you otherwise it is because you made a mistake. More studying won't help you find the mistake. I won't help you find the mistake.

It is easy to show from a spacetime diagram that in SR inertial frames the distance covered by light is always equal to the time taken. Always. Ever and ever.

Why don't you accept this ? It seems to me you won't believe it and have some reason that it must be wrong.

 

[Janus]

But let me explain what exactly I mean. suppose there is a spacecraft moving at 0.9C and at an exact moment it is far 3*10^8 meters from a space station. craft is moving away from station .At that moment the station lights a beam into the space craft. The distance between the craft and the station measured by the moving craft is 1307669683 using length contraction. and since C must be 3*10^8, then time for light beam to reach the craft must be 0.4358898944 second from craft's perspective. This means It should take 1 second from perspective of station according to time dilation, but in deed because the craft is moving away by 0.9C it should take 10 seconds for light beam to reach craft from station's perspective not 1 second. I feel confused about it

Again, your problem here is assuming that "at the same time" means the same for both the spacecraft and station.

To illustrate we add clocks to both the spaceship and station and assume that the spacecraft passed the station and at that moment they both set their clocks to 0

Now when the ship is 1 light sec from the station ( as measured by the station), the station sends its signal. At this moment, the station clock reads 1.111 sec. According to the station, the spaceship clock now reads 0.4843 sec because of time dilation. The signal takes 10 sec by the station clock to reach the spaceship, during which time the spaceship clock advances ~4.359 sec, which when added to the 0.4843 sec, means that the ship clock reads 4.843 sec when the light arrives.

Here's how things happen according to the spaceship. When the spaceship is the 0.4359 light sec away from the station( as measured by the ship) you mentioned above, its clock reads 0.4843 sec, however the station clock, due to time dilation only reads ~0.2111 secs. In other words, it hasn't yet reached the time it needs to read when it sends the signal . The ship must wait until its clock reads 2.549 before the clock on the station reads 1.111 sec and sends the signal. At which time, it will be 2.294 light seconds from the station. it will take the light 2.294 sec to travel the distance between them, meaning the ship clock reads 4.843 second when the light arrives, Which is the same answer we got according to the station.

So in order for both the ship and station to agree that the light left the station when its clock read 1.111 sec and arrived at the ship when its clock reads 4.843 sec, both the ship and station have to measure the light as traveling at c relative to themselves.

 

[Dale]

I thought X was distance moved.

No. In the Lorentz transform x is position in some inertial frame. If $x_1$ is the position at the beginning and $x_2$ is the position at the end, then $\Delta x=x_2-x_1$ is the distance moved.

 

[harrylin]

[..]

1. About synchronizing clocks with and without the observer moving.
https://www.physicsforums.com/threads/twin-paradox-and-asymmetry.816455/
You can find his clips somewhere from that thread.
I think, if an amateur like me understand (or at least grasp), perhaps you can understand more.
From his answer, I think I can understand that simultaneity events are affected by relativity.
[..]

Ah yes, you mean this one: #9
That's a very clear animation by Janus!

 

[Amr Elsayed]

So in order for both the ship and station to agree that the light left the station when its clock read 1.111 sec and arrived at the ship when its clock reads 4.843 sec, both the ship and station have to measure the light as traveling at c relative to themselves.

That's good. I got it. you exactly understood my problem and had a good way to explain it. Thanks for help, but I still have some wonders
For a laser being shot from the other direction " toward the ship" assuming our clocks were on time then I started to accelerate to reach 0.9C should I see the laser being shot after still observers do or before. I am asking because I don't know what acceleration has to do with time flow of ship clock might affect it, and I had to in order to make sure that both clocks were on time.

I think, if an amateur like me understand (or at least grasp), perhaps you can understand more.

Just don't be discourage if you don't understand

I really want to thank you, you deserve respect, but If there is an amateur here it's the guy that just is done with his first year at high school "me"
I am not discourage and I learned to be more specific while asking to get maximum help ,because If someone else understands your problem and give you what you need, you simply get it

Why don't you accept this ? It seems to me you won't believe it and have some reason that it must be wrong.

Actually, this never was the purpose of the question. I'm sure light always goes at C , but I wanted to know what nature does to keep it and it was time dilation and length contraction and simultaneity. And It did sink into some extent :)

 

[Stephanus]

Sorry, can I ask a question here?
In one of the Lorentz Transformation formula
http://en.wikipedia.org/wiki/Lorentz_transformation#Boost_in_the_x-direction
Instead of using x and t, I use x_a and t_a and instead of using x' and t' I use x_b and t_b
I use only x-axis here.
$x_b = \gamma (x_a - vt_a)$

No. In the Lorentz transform x is position in some inertial frame. If $x_1$ is the position at the beginning and $x_2$ is the position at the end, then $\Delta x=x_2-x_1$ is the distance moved.

$\text{is }x_a = \Delta x$?

 

[Stephanus]

@Amr Elsayed

I think what you are trying to do is like proving that a triangle has three sides. The speed of light is a constant in SR by definition just like a triangle has three sides by definition.
If your numerical calculations tell you otherwise it is because you made a mistake. More studying won't help you find the mistake. I won't help you find the mistake.

It is easy to show from a spacetime diagram that in SR inertial frames the distance covered by light is always equal to the time taken. Always. Ever and ever.

Why don't you accept this ? It seems to me you won't believe it and have some reason that it must be wrong.

Yeah, and it is that "ever and ever" that causes everything else to come up. Time dilation, length contraction, simultaneity of events, time machine idea, twin paradox and many things.

 

[Dale]

$\text{is }x_a = \Delta x$?

No. As you said earlier your $x_a$ is everyone else's $x$

Instead of using x and t, I use xa and ta

 

[Stephanus]

No. As you said earlier your $x_a$ is everyone else's $x$

Is X_a a coordinate or distance?
Or is X_a = X₁ and X_b = X₂ in your example?
$X_b = \gamma(x_a - vt_a)$

 

[harrylin]

[..]
For a laser being shot from the other direction " toward the ship" assuming our clocks were on time then I started to accelerate to reach 0.9C should I see the laser being shot after still observers do or before. I am asking because I don't know what acceleration has to do with time flow of ship clock might affect it, and I had to in order to make sure that both clocks were on time.

[emphasis mine]
I 'm not sure to understand your question, but still I think that I can give an answer - see next!

I wanted to know what nature does to keep it and it was time dilation and length contraction and simultaneity. And It did sink into some extent :)

Glad to hear that it is starting to sink in.

However, as I told you before, nature takes care of time dilation and length contraction, but nature does not take care of simultaneity as defined by Einstein.

Apparent simultaneity is taken care of by us, by means of clock synchronization. And if you accelerate to a different state of motion without touching your clocks, then, if you have very good clocks, you can detect a difference between the speed of light in forward and backward directions. You will have to do a new synchronization if you want to use your differently moving system as a standard reference system, in which light appears to propagate at c in all directions.

 

[Dale]

Is X_a a coordinate or distance?
Or is X_a = X₁ and X_b = X₂ in your example?
$X_b = \gamma(x_a - vt_a)$

You are the one who defined X_a so you should know! You defined x_a as the same as what everyone else uses for x, and everyone else uses x as a coordinate not a distance. So x_a is defined by you as a coordinate, not a distance.

Most people use primes (e.g. x vs x') to denote different frames. Most people use subscripts to denote different events. So most people would use x_a to denote the x coordinate of event "a". You have defined it differently, which is your prerogative, but I don't understand why you keep asking other people what is meant by the variables that you have defined. You defined it, so you should know.

x_a is NOT the same as x₁ and x_b is NOT the same as x₂. I used subscripts to denote different events, you used subscripts to denote reference frames. You yourself defined them!

 

[Stephanus]

I really want to thank you, you deserve respect, but If there is an amateur here it's the guy that just is done with his first year at high school "me"
I am not discourage and I learned to be more specific while asking to get maximum help ,because If someone else understands your problem and give you what you need, you simply get it

Oh, I just remember something
"A good theory must be able to be explained to a six years old" - Albert 1879 - 1955.
If you can't understand, perhaps you're not six years old?

 

[harrylin]

Is X_a a coordinate or distance?
Or is X_a = X₁ and X_b = X₂ in your example?
$X_b = \gamma(x_a - vt_a)$

Maybe the following helps.

Apparently you use the index a for system S, and the index b for system S'. However that will quickly be confusing because many people designate points with a, b etc. So I will use modern standard notation instead: x, t etc. relate to "stationary" system S, and x', t' etc. relate to "moving" system S'.

$x' = \gamma(x - vt)$

[EDIT:] Therefore, for points a and b on the X-axis (they are also points on the X' axis):

$x'_b = \gamma(x_b - vt_b)$
$x'_a = \gamma(x_a - vt_a)$
---------------------------------- - (subtraction)
$x'_b - x'_a = \gamma(x_b - x_a - v(t_b - t_a))$

For measurements of those points at the same time according to system S, t_b - t_a = 0.

Then we obtain for t_a=t_b:
$\Delta x' = x'_b - x'_a = \gamma(x_b - x_a)$

BTW, it is enlightening to figure out what you find for t'_a=t'_b

 

[Stephanus]

You are the one who defined X_a so you should know! You defined x_a as the same as what everyone else uses for x, and everyone else uses x as a coordinate not a distance. So x_a is defined by you as a coordinate, not a distance.

Most people use primes (e.g. x vs x') to denote different frames. Most people use subscripts to denote different events. So most people would use x_a to denote the x coordinate of event "a". You have defined it differently, which is your prerogative, but I don't understand why you keep asking other people what is meant by the variables that you have defined. You defined it, so you should know.

X_a is NOT the same as X₁ and X_b is NOT the same as X₂. I used subscripts to denote different events, you used subscripts to denote reference frames. You yourself defined them!

Okay, perhaps you would tell me what X is in Lorentz equation.
http://en.wikipedia.org/wiki/Lorentz_transformation#Boost_in_the_x-direction
I copy it literally
$t' = \gamma(t-\frac{vx}{c^2})$
$x'=\gamma(x-vt)$
What is X in this context?
Position/Coordinate?
Length?
And what is $x'=\gamma(x-vt)$ in English?
'His' position in my frame would be x' if I'm in x times gamma?

 

[Stephanus]

Apparently you use the index a for system S, and the index b for system S'

Yes, yes, that's right.
What are you? A language teacher?

Therefore, for points a and b on the X_a-axis (they are also points on the X_b axis):

$x'_b = \gamma(x_b - vt_b)$
$x'_a = \gamma(x_a - vt_a)$
---------------------------------- - (subtraction)
$x'_b - x'_a = \gamma(x_b - x_a - v(t_b - t_a))$

Thank you, thank you

 

[Stephanus]

Therefore, for points a and b on the X_a-axis (they are also points on the X_b axis):

"Therefore, for points a and b on the X-axis (they are also points on the X' axis)" in "Modern standard notation" you mean?

 

[harrylin]

"Therefore, for points a and b on the X-axis (they are also points on the X' axis)" in "Modern standard notation" you mean?

Ah thanks for spotting that - I had started answering in your notation but changed my mind. Corrected now.

 

[PWiz]

Okay, perhaps you would tell me what X is in Lorentz equation.
http://en.wikipedia.org/wiki/Lorentz_transformation#Boost_in_the_x-direction
I copy it literally
$t' = \gamma(t-\frac{vx}{c^2})$
$x'=\gamma(x-vt)$
What is X in this context?
Position/Coordinate?
Length?
And what is $x'=\gamma(x-vt)$ in English?
'His' position in my frame would be x' if I'm in x times gamma?

x is the coordinate in S frame, whereas x' is the coordinate in S' frame (NOT distance). Your final equation enables you to find the position coordinate in frame S' (which is denoted by x') using the position coordinate and time from frame S (which are denoted by x and t). It roughly the answer to the question "if that ball is recorded to have position coordinate x at time t in frame S, what will be that ball's position coordinate in frame S' ?".

 

[Stephanus]

x is the coordinate in S frame, whereas x' is the coordinate in S' frame (NOT distance). Your final equation enables you to find the position coordinate in frame S' (which is denoted by x') using the position coordinate and time from frame S (which are denoted by x and t). It roughly the answer to the question "if that ball is recorded to have position coordinate x at time t in frame S, what will be that ball's position coordinate in frame S' ?".

Thank you, thank you.

 

[Amr Elsayed]

I 'm not sure to understand your question, but still I think that I can give an answer - see next!

I wanted to see how simultaneity works in the other case " light goes toward the moving ship" . Briefly I was asking if I shall also see the light emission after still observers do knowing that they were as distant as me when light was emitted . If our clocks were on time before and due to time dilation I then should see light emission after they do since light emission is related to a specific time at the still clock of station . I'm not sure If my synchronization to set clocks on time first is right and I'm not sure if should see light before still observers does or after

Glad to hear that it is starting to sink in.
However, as I told you before, nature takes care of time dilation and length contraction, but nature does not take care of simultaneity as defined by Einstein.

Thank you
I agree with you that we take care of simultaneity

Oh, I just remember something
"A good theory must be able to be explained to a six years old" - Albert 1879 - 1955.
If you can't understand, perhaps you're not six years old?

Yeah Good point

 

[Stephanus]

This is 'Janus' train
Let
A1: Front Mark
A2: Bow of the Train, Front
B1: Rear Mark
B2: Aft of the Train, Back
M: Mid point between A1 and B1
D: Distance between A1 and M, let say D = 1.6 Light second
E: Distance between A1 and B1 = 2 * D = 3.2 lt
V: Velocity of the railway = 0.6c
L: is the length of the train = ...??
Yes! The railway is moving. The train stops. I don't know if this makes sense in real world.

Everything is in Train Frame.
Okay...
WHEN A2 meets A1, the train bow flashes a signal
Let V_a = c-V = 0.4c
T_a = D/V_a = 4 sec
The signal will reach M in 4 sec

WHEN B2 meets B1, the train aft flashes a signal
Let V_b = c+V = 1.6c
T_b = D/V_b = 1 sec

So B2 shouldn't meet B1 at the same time A2 meets A1, right? There's simultaneity different here
In fact B2 should 'wait' for 3 seconds or the length of the train should be longer than A1-B1, there's length contraction here
Additional train length is 3 seconds * V
So
L = 3 seconds * V + E
$L = (T_a - T_b)V + E$

$L = (\frac{D}{c-V}-\frac{D}{c+V})*V+E$

$L = (\frac{Dc+DV-(Dc-DV)}{c^2-V^2})*V+E$

$L = (\frac{2DV}{c^2-V^2})*V+E$

$E = 2D$, so

$L = (\frac{EV^2}{c^2-V^2})+E$

If we substitute C with 1 and V is a factor of C, so

$L = (\frac{EV^2}{1-V^2})+E$

$\frac{L}{E} = (\frac{V^2}{1-V^2})+1$
$\frac{L}{E} = (\frac{1}{1-V^2})$
So, the railway length is 3.2 lt and the train length is 5 lt for V = 0.6c
That way, 5 lt - 3.2 lt =1.8 lt
If V = 0.6 then, it takes 3 seconds for the back of the train to reach 1.8 lt. And flashes a signal which hit M at the same time the bow signal hits M
It makes sense,
But...
Why
$\gamma = \frac{1}{\sqrt{1-V^2}}$?
not this one
$\text{ratio} = \frac{1}{1-V^2}$
Do I mistakenly make the equation?
But if Lorentz is right, WHICH I KNOW HE IS!, the length of the train will be 4 lt not 5 lt. Doesn't make sense in real world. Or it does?

 

[Stephanus]

Why
$\gamma = \frac{1}{\sqrt{1-V^2}}$?
not this one
$\text{ratio} = \frac{1}{1-V^2}$
Do I mistakenly make the equation?
But if Lorentz is right, WHICH I KNOW HE IS!, the length of the train will be 4 lt not 5 lt. Doesn't make sense in real world. Or it does?

Ahhh, it works both ways. The train and the railway.
The length of the train AT REST is 5 light seconds.
And the length of the railway AT REST is 4 ls.
The ratio doesn't have to be
$\frac{1}{1-V^2}$
But,
$\frac{1}{\sqrt{1-V^2}}$ is enough.
The length of the moving rail is not computed from the train, but from the railway AT REST.
So $4\text{ ls } * \frac{1}{\sqrt{1-v^2}} \text { is } 3.2 \text{ls}$
And for the moving train...
$5 \text { ls } * \frac{1}{\sqrt{1-v^2}} \text { is } 4 \text{ ls}$
I had this answer when I drove at highway watching road milestone, coming back from my father in law house at the country.
But, it's been 24 hours since I post this posting. Why nobody answers?

 

[harrylin]

Ahhh, it works both ways. The train and the railway.
[..]
I had this answer when I drove at highway watching road milestone, coming back from my father in law house at the country.
But, it's been 24 hours since I post this posting. Why nobody answers?

There is a limited amount of people here, with limited free time, and too many topics. I didn't understand your question, and also I don't understand your new insight, except that it looks as if you now for the first time understand that time dilation is mutual. I'm glad that you apparently got it now.

 

[Stephanus]

There is a limited amount of people here, with limited free time, and too many topics. I didn't understand your question, and also I don't understand your new insight, except that it looks as if you now for the first time understand that time dilation is mutual. I'm glad that you apparently got it now.

Sorry, I didn't mean to rush everybody with my (endless) questions .
It's just a rhetoric statement.
Actually I want to know about twins paradox and why there's twins paradox but the universe HAS NO frame of reference.
But to get there, I've been asking about
Motion in Space
Doppler
Lorentz
Relativity
Signal
etc...
But now I know I'm on the right track, but not there, yet.
I want to thank you personally HarryLin for your helps to me all this time.

 

[Ibix]

There is a twin paradox because elapsed time is the "length" of a path through space-time, and the twins take different paths. There's not really any more mystery to it than that.

 

[Stephanus]

There is a twin paradox because elapsed time is the "length" of a path through space-time, and the twins take different paths. There's not really any more mystery to it than that.

Thanks Ibix for your answer.
But I think, it's still far away from my destination.
Now, I just understand how Hendrik discovered this formula.
$\gamma = \frac{1}{\sqrt{1-v^2}}$
I stil want to study why
$t' = \gamma(t-\frac{vx}{c^2})$ and
$x' = \gamma(x-vt)$
through train/platform simulation, by myself just like
https://www.physicsforums.com/threads/length-contraction.817911/#post-5137268And after that I'll fire my (endless) questions (again).